quiz5_solution

2 x 2t2 3 and y t4 find dy d2 y and 2 when t 1 dx dx

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Unformatted text preview: v (t)|, the answer is |v (2)| = 3m/sec. 2. x = 2t2 + 3 and y = t4 . Find dy d2 y and 2 when t = −1. dx dx dy = dx dy dt dx dt = 4t3 = t2 4t dy Therefore, |t=−1 = 1. dx Next, d2 y d dy = ( )= dx2 dx dx d dy () dt dx dx dt = 2t 1 =. 4t 2 d2 y 1 Therefore, 2 |t=−1 = . dx 2 3. Find the equation of the line normal to the curve x2 y 2 = 9 at the point (−1, 3). By implicit differentiation and the Product rule, dy =0 dx dy x2 2 y = −2xy 2 dx dy −2xy 2 =2 . dx x 2y 2xy 2 + x2 2y dy Therefore, the slope of the line tangent to the curve at (−1, 3) is |(−1,3) = 3. We are, dx however, looking for the equation of the line normal to the curve at this point, so the 1 2 1 slope of this line will be − , and the normal line equation is 3 1 y − 3 = − (x + 1). 3...
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